Work required to increase the radius of a sphere by dr

  • #1

Homework Statement


To assemble a uniformly charged sphere, assemble it like a snowball, layer by layer, each time bringing in an infinitesimal charge dq from far away and smearing it uniformly over the surface, thereby increasing the radius. How much work dW does it take to build up the radius by an amount dr? Integrate this to find the work necessary to create the entire sphere of radius R and total charge q.


Homework Equations



W = 1/2∫ρVdτ

The Attempt at a Solution



dW = k*q*(dq/R)

This is the work needed to add a charge dq from infinity to the outer shell of the sphere

it will then increase it by dr based on the charge density added to the sphere.

so dq = 4πr^2*ρ*q*dr

making dW = (k*q*4πr^2*ρ*dr)/R

So W = k*q*4πr^2*∫ρ r^2 dr dτ

I am confused if i have set this up right and on what interval i need to integrate.
 
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Answers and Replies

  • #2
For your final expression for dW, write out in words what it means.
It is the work needed to...
 
  • #3
dW is needed to take a charge from infinitely far away and distribute it even across the surface of the sphere thus increasing the radius by an infinitesimally small amount.

Thus making W to make the sphere be the the sum of dW from 0 to R which answers my integration from where problem. Thank you for that.

I am still confused on if i have set up the problem right though. So would it be correct to just integrate it so that

W = k*q*4πr^2*ρ* ∫ r^2 dr from 0 to R?
 
  • #4
dW = k q1 q2 /r in general
Here, q1 is the charge q as a function of r. What is that? Hint: charge = charge density x volume.
And here q2 = dq.
So express dW as a function of r and dq: dW = k q(r) dq/r
Then, take dq = f(r) dr and substitute so that dW does not have q in it. You have eliminated q and dq in your equation for dW. Then integrate from r = 0 to r = R.

And whatever power of r you wind up with, it all has to go INSIDE the integral sign!
 
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  • #5
dW = k*(dq/R)

This is the work needed to add a charge dq from infinity to the outer shell of the sphere
I'm thinking - doesn't the amount of work you have to do also depend on how much charge has already been moved? Or did you take account of that later?

so dq = 4πr^2*ρ*dr
... dq would be the amount of charge in a spherical shell thickness dr?
i.e. the surface area of the underlying volume radius r times the thickness. That how you did it?

You've already seen that writing down your reasoning is good for sorting out your thoughts later. It is also good for getting those elusive final marks on long-answer questions.
 
  • #6
Thank you, i was able to the problem out.
 
  • #7
Well done - want to share for those who google here with the same question?
 
  • #8
Yes, I would like to compare my answers with yours.
 

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